Prove: (1-tan A)^2+(1-cot A)^2=(sec A-cosec A)^2 | Trig Identity

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Hey guys! Today, we're diving deep into the fascinating world of trigonometry to tackle a pretty neat identity. We're going to prove that $(1-\tan A)^2+(1-\cot A)^2$ is indeed equal to $(\sec A-\operatorname{cosec} A)^2$. Buckle up, because we're about to break this down step by step, making sure everyone, whether you're a trig whiz or just starting out, can follow along. This isn't just about memorizing formulas; it's about understanding the underlying principles and how these trigonometric functions relate to each other. So, let’s jump right in and unravel this mathematical puzzle together! Remember, the beauty of math lies in its logical progression, so we'll take it slow and steady, ensuring every step makes perfect sense.

Breaking Down the Left-Hand Side (LHS)

Okay, let's start with the left-hand side (LHS) of our equation: $(1-\tan A)^2+(1-\cot A)^2$. Our mission here is to simplify this expression, and the best way to do that is by expanding those squares and seeing where the trigonometric magic takes us. Remember, we're not just blindly applying formulas; we're strategically manipulating the expression to reveal its hidden form. Think of it like a mathematical treasure hunt, where each step brings us closer to the final solution. So, grab your mathematical shovels, and let's dig in!

First up, let's expand $(1-\tan A)^2$. This is a classic algebraic expansion: $(a-b)^2 = a^2 - 2ab + b^2$. Applying this, we get:

$(1-\tan A)^2 = 1^2 - 2(1)(\tan A) + \tan^2 A = 1 - 2\tan A + \tan^2 A$

Easy peasy, right? Now, let's move on to the second term, $(1-\cot A)^2$. We'll use the same expansion formula here:

$(1-\cot A)^2 = 1^2 - 2(1)(\cot A) + \cot^2 A = 1 - 2\cot A + \cot^2 A$

Great! Now we have both expanded forms. Let’s combine them to get the full expansion of the LHS:

$(1-\tan A)^2+(1-\cot A)^2 = (1 - 2\tan A + \tan^2 A) + (1 - 2\cot A + \cot^2 A)$

Now, let’s group the like terms together. We'll bring the constants together, the terms with tangents and cotangents, and the squared terms. This is like sorting your LEGO bricks before you start building – it makes the whole process much smoother.

$= 1 + 1 - 2\tan A - 2\cot A + \tan^2 A + \cot^2 A$

Combining the constants, we simplify to:

$= 2 - 2\tan A - 2\cot A + \tan^2 A + \cot^2 A$

Now, we're at a crucial juncture. We have an expanded form, but it doesn't immediately scream the right-hand side. This is where our trigonometric identities come into play. We need to massage this expression, using our knowledge of trig identities, to transform it into something that resembles $(\sec A-\operatorname{cosec} A)^2$. Remember, the goal isn't just to get the answer; it's to understand how we get there. So, let’s keep pushing forward, armed with our trig toolkit!

We can rewrite the expression as:

$2 - 2\tan A - 2\cot A + \tan^2 A + \cot^2 A$

This is our simplified LHS. Now, let’s keep this aside and move on to the RHS, where we’ll see if we can make it match this expression. This is the exciting part where we start to see connections and the puzzle pieces begin to fit together. Stick with me, guys; we're getting there!

Simplifying the Right-Hand Side (RHS)

Alright, let's shift our focus to the right-hand side (RHS) of the identity: $(\sec A-\operatorname{cosec} A)^2$. Just like we did with the LHS, our initial strategy here is to expand this expression. Expanding squares is a fundamental technique in algebra, and it’s going to be our trusty tool once again. Remember, the key to simplifying complex expressions is often breaking them down into smaller, more manageable parts. So, let's roll up our sleeves and get to work on this RHS!

Using the same expansion formula $(a-b)^2 = a^2 - 2ab + b^2$, we can expand $(\sec A-\operatorname{cosec} A)^2$ as follows:

$(\sec A-\operatorname{cosec} A)^2 = \sec^2 A - 2(\sec A)(\operatorname{cosec} A) + \operatorname{cosec}^2 A$

Okay, that looks promising! We've got squares of secant and cosecant, and a term involving both. But we're not done yet. Now, we need to dig a little deeper and see how we can further simplify this expression. This is where our knowledge of fundamental trigonometric identities comes into play. We need to think about how secant and cosecant relate to other trig functions like sine and cosine. Remember, these identities are the building blocks of trigonometric manipulations, and mastering them is crucial for solving problems like this.

Recall that $\sec A = \frac{1}{\cos A}$ and $\operatorname{cosec} A = \frac{1}{\sin A}$. This is a crucial step because it allows us to rewrite the expression in terms of sine and cosine, which are often easier to work with. Let’s substitute these into our expanded RHS:

$\sec^2 A - 2(\sec A)(\operatorname{cosec} A) + \operatorname{cosec}^2 A = \frac{1}{\cos^2 A} - 2\left(\frac{1}{\cos A}\right)\left(\frac{1}{\sin A}\right) + \frac{1}{\sin^2 A}$

Now, we have an expression in terms of sines and cosines. Let’s simplify the middle term:

$= \frac{1}{\cos^2 A} - \frac{2}{\sin A \cos A} + \frac{1}{\sin^2 A}$

This is starting to look interesting! We have three terms, each involving sine and/or cosine. To combine these terms, we need to find a common denominator. The common denominator here is $\sin^2 A \cos^2 A$. So, let’s rewrite each term with this denominator:

$= \frac{\sin^2 A}{\sin^2 A \cos^2 A} - \frac{2\sin A \cos A}{\sin^2 A \cos^2 A} + \frac{\cos^2 A}{\sin^2 A \cos^2 A}$

Now that we have a common denominator, we can combine the numerators:

$= \frac{\sin^2 A - 2\sin A \cos A + \cos^2 A}{\sin^2 A \cos^2 A}$

Hey, look at that numerator! It’s a familiar form: $\sin^2 A + \cos^2 A - 2\sin A \cos A$. We know that $\sin^2 A + \cos^2 A = 1$, a fundamental trigonometric identity. Let’s substitute that in:

$= \frac{1 - 2\sin A \cos A}{\sin^2 A \cos^2 A}$

We're making excellent progress! We've simplified the RHS to a single fraction. But how does this relate back to our simplified LHS? This is where we need to start thinking strategically about how to bridge the gap between the two sides. We might need to manipulate this expression further, or we might need to revisit our simplified LHS and see if we missed anything. Remember, in math, sometimes you need to try a few different paths before you find the right one. So, let’s keep our eyes peeled for connections and see where this takes us!

Bridging the Gap: Connecting LHS and RHS

Okay, guys, we've got our simplified LHS: $2 - 2\tan A - 2\cot A + \tan^2 A + \cot^2 A$, and our simplified RHS: $\frac{1 - 2\sin A \cos A}{\sin^2 A \cos^2 A}$. Now comes the exciting part – bridging the gap between these two expressions! This is where we put on our detective hats and look for clues, connections, and pathways that will lead us to the final proof. It’s like solving a puzzle where we have two separate pieces, and our goal is to make them fit perfectly together. So, let's sharpen our minds and get ready to connect the dots!

Looking at the LHS, we have terms involving tangents and cotangents. On the RHS, we have an expression in terms of sines and cosines. A natural strategy here is to convert the tangents and cotangents in the LHS to sines and cosines. This will allow us to compare the two sides more directly. Remember, converting to sines and cosines is a powerful technique in trigonometry because it allows us to use the fundamental identities and relationships between these functions. So, let’s apply this strategy and see where it takes us.

Recall that $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$. Let’s substitute these into the LHS:

$2 - 2\tan A - 2\cot A + \tan^2 A + \cot^2 A = 2 - 2\left(\frac{\sin A}{\cos A}\right) - 2\left(\frac{\cos A}{\sin A}\right) + \left(\frac{\sin A}{\cos A}\right)^2 + \left(\frac{\cos A}{\sin A}\right)^2$

Now, let’s simplify the squares:

$= 2 - 2\frac{\sin A}{\cos A} - 2\frac{\cos A}{\sin A} + \frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\sin^2 A}$

We need to combine these terms, so let’s find a common denominator. The common denominator for all these terms is $\sin^2 A \cos^2 A$. Let’s rewrite each term with this denominator:

$= \frac{2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} - \frac{2\sin^3 A \cos A}{\sin^2 A \cos^2 A} - \frac{2\sin A \cos^3 A}{\sin^2 A \cos^2 A} + \frac{\sin^4 A}{\sin^2 A \cos^2 A} + \frac{\cos^4 A}{\sin^2 A \cos^2 A}$

Now we can combine the numerators:

$= \frac{2\sin^2 A \cos^2 A - 2\sin^3 A \cos A - 2\sin A \cos^3 A + \sin^4 A + \cos^4 A}{\sin^2 A \cos^2 A}$

This looks a bit intimidating, but don't worry! We can simplify this further. Let’s focus on the numerator. We can factor out a $-2\sin A \cos A$ from the second and third terms:

$= \frac{2\sin^2 A \cos^2 A - 2\sin A \cos A(\sin^2 A + \cos^2 A) + \sin^4 A + \cos^4 A}{\sin^2 A \cos^2 A}$

We know that $\sin^2 A + \cos^2 A = 1$, so let’s substitute that in:

$= \frac{2\sin^2 A \cos^2 A - 2\sin A \cos A + \sin^4 A + \cos^4 A}{\sin^2 A \cos^2 A}$

Now, let’s rearrange the terms in the numerator:

$= \frac{\sin^4 A + \cos^4 A + 2\sin^2 A \cos^2 A - 2\sin A \cos A}{\sin^2 A \cos^2 A}$

Notice that $\sin^4 A + \cos^4 A + 2\sin^2 A \cos^2 A$ looks like a perfect square! In fact, it’s $(\sin^2 A + \cos^2 A)^2$. And we know that $\sin^2 A + \cos^2 A = 1$, so this whole term simplifies to $1^2 = 1$. Let’s substitute that in:

$= \frac{1 - 2\sin A \cos A}{\sin^2 A \cos^2 A}$

Wait a minute… This looks familiar! It’s exactly the same as our simplified RHS! We’ve done it! We’ve successfully transformed the LHS into the RHS. This is a moment of mathematical triumph! We started with two seemingly different expressions, and through careful manipulation and application of trigonometric identities, we’ve shown that they are indeed equal. Give yourselves a pat on the back, guys; we’ve conquered this trigonometric mountain together!

Conclusion: The Identity Stands Proven

Alright, guys, let's take a moment to celebrate! We’ve successfully proven the trigonometric identity $(1-\tan A)^2+(1-\cot A)^2=(\sec A-\operatorname{cosec} A)^2$. We journeyed through expanding squares, converting between trigonometric functions, finding common denominators, and applying key identities. It was a challenging but rewarding process, and we emerged victorious! This wasn't just about memorizing steps; it was about understanding the underlying principles and how different trigonometric functions relate to each other. Remember, math isn’t just about getting the right answer; it’s about the journey of discovery and the satisfaction of solving a complex problem.

We started by breaking down the left-hand side (LHS), expanding the squares, and simplifying the expression. Then, we tackled the right-hand side (RHS), using our knowledge of secant and cosecant to rewrite it in terms of sine and cosine. The crucial step was recognizing the connection between the simplified LHS and RHS. By converting the tangents and cotangents on the LHS to sines and cosines, we were able to transform it into the exact same form as the RHS. This demonstrated the equality of the two sides and completed our proof.

This identity is a beautiful example of the interconnectedness of trigonometric functions. It showcases how seemingly different expressions can be equivalent through the clever application of trigonometric identities. Mastering these identities and techniques is essential for anyone delving deeper into trigonometry and calculus. So, keep practicing, keep exploring, and keep challenging yourselves with new mathematical puzzles. The more you engage with these concepts, the more intuitive they will become.

Remember, the key to success in mathematics is not just memorization, but understanding. Focus on the “why” behind each step, and the “how” will naturally follow. And most importantly, don’t be afraid to make mistakes! Mistakes are valuable learning opportunities. They help us identify areas where we need to strengthen our understanding and guide us towards more effective problem-solving strategies.

So, congratulations again on conquering this trigonometric challenge! You've shown yourselves that you have the skills and the perseverance to tackle complex problems. Keep up the great work, and I look forward to exploring more mathematical adventures with you guys in the future!